Metric Space is Lindelöf iff Second-Countable

Theorem

Let $M = \struct {A, d}$ be a metric space.

Then $M$ is Lindelöf if and only if $M$ is second-countable.


Proof

Sufficient Condition

We have from Second-Countable Space is Lindelöf that second-countability implies Lindelöf in all topological spaces, regardless of whether they are metric spaces or not.

$\Box$


Necessary Condition

Suppose $M = \struct {A, d}$ is Lindelöf.

Let us define the open covers on $A$:

$\CC_k = \set {\map {N_{1/k} } x: x \in S}$

for all $k \in \N_{>0}$.

As $M$ is Lindelöf, each one of these has a countable subcover.

The union of all these subcovers is a countable basis for the topology on $A$.

Hence the result, by definition of second-countable space.

$\blacksquare$


Sources

  • 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
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